Proof: Every convergent sequence of real numbers is bounded 
(Taken from an old exam - no homework:)
Prove that every convergent sequence of real numbers is bounded.

It would be cruel if our professor would take a task like that in the exam because I rather prefer calculating than doing theory (in an exam where you haven't got much time to describe things..). But I'll be honest and say I dislike theory in general! :-)
Anyway, for the task, I really got no idea how to proof that but I'll give it a try...

We know that if a sequence is convergent it has exactly ONE limit where the sequences (sry my English here) accumulate. So the sequence will have an upper bound or lower bound, based on the limit, thus it will be surely bounded if it's convergent sequence.

Even if my proof is wrong (most likely..) I'd like to see a proof like mine, where you avoid using (complicated) maths symbols. Just using words any human can understand would be awesome! :-)
Of course a more formal proof is welcome too but not necessarily preferred ^.^
 A: Okay, let's start off with the definition of a convergent sequence:
$$\lim_{n \to \infty} a_{n}=a  \Leftrightarrow (\forall \varepsilon>0)(\exists n_{0} \in \mathbb{N})(\forall n > n_{0}) |a_{n}-a|<\varepsilon$$
What this definition states is that after a certain term of a sequence, all the following terms will be close to the limit.
Now, we can take an arbitrary $\varepsilon >0$ and for that $\varepsilon$ we know that all the terms of the sequence whose index is greater than $n_{0}(\varepsilon)$ are in the interval $(a-\varepsilon , a+\varepsilon )$. So almost all of the sequence is bounded. What about $a_{0}, a_{1}, ... , a_{n_{0}}$? Well, there's a finite number of them, so we can choose a minimal and a maximal element, and denote them as $A$ and $B$, respectively.
Now, it's apparent that the sequence is bounded, because all of its terms are in the interval $[\min\{A, a-\varepsilon \}, \max\{B, a+\varepsilon\}]$.
I hope the proof was clear enough.
A: The basic idea is:


*

*Any finite set of real numbers is bounded.

*The tail of the sequence is bounded by the limit plus/minus epsilon from some point onwards.


So you can divide it into a finite set of the first say $N-1$ elements of the sequence and a bounded set of the tail from $N$ onwards. Each of those will be bounded by 1. and 2. above. The conclusion follows.

If this helps, perhaps you could even show the effort to rephrase this approach into a formal proof forcing yourself to apply the proper mathematical language with epsilon-delta definitions and all that? Post it as an answer to your own question ...
